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Svet_ta [14]
4 years ago
6

What is the difference? X/x2+3x+2 - 1/(X+ 2)(x+1)

Mathematics
1 answer:
Anarel [89]4 years ago
5 0

Answer:

D

Step-by-step explanation:

\[\frac{x}{x^{2} +3x+2} -\frac{1}{(x+2)(x+1)} =\frac{x}{x^2+3x+2} -\frac{1}{x(x+1)+2(x+1)} =\frac{x}{x^{2}+3x+2 } -\frac{1}{x^{2} +2x+x+2} =\frac{x}{x^{2} +3x+2} -\frac{1}{x^{2} +3x+2} =\frac{x-1}{x^{2} +3x+2} \]

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Find the length of an arc that has central angle, 50 degrees, and radius, 3.
kotegsom [21]
First we find the circumference
c=2pir
r=3
2pi3
6pi=c

then find out how much of the circle is 50 degreese
50/360=5/36
6pi times 5/36=5pi/6=aprox 2.62

it should be 5pi/6

8 0
4 years ago
Solve this equation for<br> a. A ÷ 2 = 4<br> A. 8<br> B. 6<br> C. 4<br> D. 2
Volgvan
A ÷ 2 = 4

Multiply both sides by 2.

a ÷ 2 × 2= 4 × 2

a = 4 × 2

a = 8

Your final answer is A. 8.
6 0
3 years ago
Please answer this question
Tanya [424]
The answer is about 1.6
7 0
4 years ago
Which answer choice shows 21.97 written in expanded form? A. 2 + 1 + 9 + 7 B. 20 + 1 + 0.9 + 0.7 C. 20 + 10 + 0.9 + 0.07 D. 20 +
Simora [160]

Answer:

20 + 1 + 0.9 + 0.07

Step-by-step explanation:

Given the number 21.97 ;

To wite in expanded form:

Tens ___ Unit ____ tenth ____ hundredth

2_______ 1 _______ 9_________ 7

2 = 20

1 = 1

9 = 0.9

7 = 0.07

Hence,

20 + 1 + 0.9 + 0.07

6 0
3 years ago
(a) The plane y + z = 13 intersects the cylinder x2 + y2 = 25 in an ellipse. Find parametric equations for the tangent line to t
klemol [59]

Answer:

Step-by-step explanation:

We have a curve (an ellipse) written as the system of equations

\begin{cases} y+z &= 13\\ x^2+y^2 &= 25\end{cases}.

And we want to calculate the tangent at the point (3,4,9).

The idea in this problem is to consider two variables as functions of the third. Usually we consider y and z as functions of x. Recall that a curve in the space can be written in parametric form in terms of only one variable. In this case we are considering the ‘‘natural’’ parametrization (x, y(x), z(x)).

Recall that the parametric equation of a line has the form

r(t)=\begin{cases} x(t) &= x_0 + v_1t \\ y(t) &= y_0 +v_2t\\ z(t) &= z_0 +v_3t \end{cases},

where (x_0,y_0,z_0) is a point on the line (in this particular case is (3,4,9)) and (v_1,v_2,v_3) is the direction vector of the line. In this case, the direction vector of the line is the tangent vector of the ellipse at the point (3,4,9).

Now, if we have the parametric equation of a curve (x, y(x), z(x)) its tangent line will have direction vector (1, y'(x), z'(x)). So, as we need to calculate the equation of the tangent line at the point (3,4,9) = (3, y(3), z(3)), we must obtain the tangent vector (1, y'(3), z'(3)). This part can be done taking implicit derivatives in the systems that defines the ellipse.

So, let us write the system as

\begin{cases} y(x)+z(x) &= 13\\ x^2+y^2(x) &= 25\end{cases}.

Then, taking implicit derivatives:

\begin{cases} y'(x)+z'(x) &= 0 \\ 2x+2y(x)y'(x) &= 0\end{cases}.

Now we substitute the values x=3 and y(3)=4, and we get the system of linear equations

\begin{cases} y'(3)+z'(3) &= 0 \\ 2\cdot 3+2\cdot 4y'(x) &= 0\end{cases},

where the unknowns are y'(3) and z'(3).

The system is

\begin{cases} y'(3)+z'(3) &= 0 \\ 6+8y'(x) &= 0\end{cases},

and its solutions are

y'(3) = -\frac{3}{4} and z'(3) = \frac{3}{4}.

Then, the direction vector of the tangent is

(1, -\frac{3}{4}, -\frac{3}{4}).

Finally, the tangent line has parametric equation

r(t)=\begin{cases} x(t) &= 3 + t \\ y(t) &= 4 -\frac{3}{4}t\\ z(t) &= 9 +\frac{3}{4}t \end{cases}

where t\in\mathbb{R}.

7 0
4 years ago
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